# Project on trigonometry

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its a good project .. office 7 ?

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Thank you very much. Your presentation has made me half of the lesson understandable. Thankx a lot

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## Presentation Transcript

### PROJECT ON TRIGONOMETRY :

PROJECT ON TRIGONOMETRY DESIGNED BY :- SHUBHAM KUMAR 10TH D ROLL NO :- 6 Sin θ Cot θ Cos θ Cosecant θ ∟A ½=θ Next Slide

### Acknowledgement :

Acknowledgement I made this project under the guidance of my mathematics teacher Ms. Jyoti Gupta mam Next Slide Previous Slide HOME

### Welcome to the world of trigonometry :

Welcome to the world of trigonometry Next Slide Previous Slide HOME

### Introduction to trigonometry :

Introduction to trigonometry The word trigonometry is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides’ ) and ‘metron’ (meaning measure). In fact, Trigonometry is the study of the relationships between the sides and angles of a triangle. Trigonometric ratios of an angle are some ratios of the sides of a right triangle with respect to its acute angles. Trigonometric identities are some trigonometric ratios for some specific angles and some identities involving these ratios. Next Slide Previous Slide HOME

### example :

example Suppose the students of a school are visiting Eiffel tower . Now, if a student is looking at the top of the tower, a right triangle can be imagined to be made as shown in figure. Can the student find out the height of the tower, without actually measuring it? Yes the student can find the height of the tower with the help of trigonometry. Next Slide Previous Slide HOME

### Trigonometric ratios :

Trigonometric ratios Let us take a right angle ABC as shown in figure. Here, ∟CAB or ∟A is an acute angle. Note the position of side BC with respect to ∟A. It faces ∟A. we call it the side opposite to ∟A(perpendicular). AC is hypotenuse of the right angle and the side AB is a part of ∟A. so, we call it the side adjacent to ∟A(base). Next Slide Previous Slide HOME

### Slide 7:

Trigonometric ratios Let us take a right angle ABC as shown in figure. Here, ∟ACB or ∟C is an acute angle. Note the position of side AB with respect to ∟C. It faces ∟C. we call it the side opposite to ∟C(perpendicular). AC is hypotenuse of the right angle and the side BC is a part of ∟C. so, we call it the side adjacent to ∟C(base). Next Slide Previous Slide HOME

### names of trigonometric ratios :

names of trigonometric ratios Next Slide Previous Slide HOME

### Trigometric ratios :

Trigometric ratios Definitions of sine cosine & tangent Next Slide Previous Slide HOME

### Triangle terminology :

Triangle terminology Opposite side Adjacent side Hypotenuse Next Slide Previous Slide HOME

### Opposite side: the side opposite the angle :

angle angle angle Opposite side: the side opposite the angle opposite opposite opposite opposite angle Opposite side Adjacent side Hypotenuse Next Slide Previous Slide HOME

### Hypotenuse: the longest side :

Hypotenuse: the longest side hypotenuse hypotenuse hypotenuse hypotenuse Opposite side Adjacent side Hypotenuse Next Slide Previous Slide HOME

### Definitions :

Definitions Tangent Sine Cosine A B 90o C a c b Next Slide Previous Slide HOME

### Tangent of angle A :

clear Sine Cosine A B 90o C a c b tan(A) = tan(B) Tangent of angle A opposite adjacent a b = Next Slide Previous Slide HOME

### Tangent of angle B :

Tangent of angle B clear Sine Cosine A B 90o C a c b tan(B) = tan(A) opposite adjacent b a = Next Slide Previous Slide HOME

### Sine of angle A :

Sine of angle A Tangent Cosine A B 90o C a c b sin(A) = = a c sin(B) opposite hypotenuse Clear Next Slide Previous Slide HOME

### Sine of angle B :

Sine of angle B Tangent Clear A B 90o C a c b sin(B) = = b c sin(A) opposite hypotenuse Cosine Next Slide Previous Slide HOME

### Cosine of angle A :

Cosine of angle A Tangent Sine Clear A B 90o C a c b cos(A) = = b c cos(B) adjacent hypotenuse A B 90o Next Slide Previous Slide HOME

### Cosine of angle B :

Cosine of angle B Tangent Sine Clear A B 90o C a c b cos(B) = = a c cos(A) adjacent hypotenuse Next Slide Previous Slide HOME

### The trigonometric ratios of the ∟A in right angle are termed as follows :- :

The trigonometric ratios of the ∟A in right angle are termed as follows :- Sin A = Perpendicular/ Hypotenuse = BC / AC Cos A = Base / Hypotenuse = AB / AC Tan A = Perpendicular / Base = BC /AB Cosec A = Hypotenuse/ Perpendicular = AC / BC Sec A = Hypotenuse / Base = AC / AB Cot A = Base / Perpendicular = AB / BC Next Slide Previous Slide HOME

### Way to learn trigonometric ratios :

Way to learn trigonometric ratios Sona Chandi Tawle Pundit Badri Prasad Har Har Bole INFORMATION S – Sin θ C – Cos θ T – Tan θ P – Perpendicular B – Base H – Hypotenuse S = P/ H C = B / H T = P / B Next Slide Previous Slide HOME

### Reciprocals of sin , cos & tan :

Reciprocals of sin , cos & tan Sin θ = reciprocal= Cosec θ Cos θ = reciprocal = Sec θ Tan θ = reciprocal = Cot θ Means :- Sin θ = 1/ Cosec θ (sin θ * cosec θ = 1 ) Cos θ = 1/ Sec θ ( cos θ * sec θ = 1 ) Tan θ = 1/ Cot θ ( tan θ * cot θ = 1 ) Next Slide Previous Slide HOME

### questions related to above topics :

questions related to above topics Calculating the value of other trigonometric ratios, if one is given. Proving type. Evaluating by using the given trigonometric ratio’s value. Next Slide Previous Slide HOME

### Type 1 – calculating value of other trigonometric ratios, if one is given. :

Type 1 – calculating value of other trigonometric ratios, if one is given. If Sin A = 3 / 4 , calculate Cos A and Tan A . Solution - Sin A = P / H = BC / AC = 3 / 4 Let BC = 3K AND , AC = 4K THEREFORE, By Pythagoras Theorem, (AB)² = (AC)² – (BC)² (AB)² = (4K)² - (3K)² AB = √7K Cos A = B / H= AB / AC = √7K / 4K = √7 / 4 Tan A = P / B = BC / AB = 3K / √7K = 3 / √7 Next Slide Previous Slide HOME

### Type 2 – proving type :

Type 2 – proving type If ∟A and ∟B are acute angles such that Cos A = Cos B, then show that ∟A = ∟B Solution - Since, Cos A = Cos B AC / AB = BC / AB therefore, AC = BC. ∟B = ∟A (angles opposite to equal sides ) Therefore , ∟A = ∟B Next Slide Previous Slide HOME

### Type 3 – evaluating by putting the given trigonometric ratio’s value :

Type 3 – evaluating by putting the given trigonometric ratio’s value If Sec A = 5 / 4 , evaluate 1 – Tan A . 1 + Tan A Solution – Sec A = H / B =AC / AB = 5 / 4 Let AC / AB = 5K / 4K. By Pythagoras Theorem , (BC)² = (AC ) ² – (AB) ² Therefore, BC = 3K So, Tan A = P / B = BC / AB = 3K / 4K = 3 / 4 1 – Tan A = 1 – 3 / 4 = 1 / 4 = 1 1 + Tan A 1 + 3 / 4 7 / 4 7 Next Slide Previous Slide HOME

### Values Of Trigonometric Ratios :

Values Of Trigonometric Ratios Next Slide Previous Slide HOME

### examples ON VALUES OF TRIGONOMETRIC RATIOS :

examples ON VALUES OF TRIGONOMETRIC RATIOS Evaluation Finding values of A and B. Next Slide Previous Slide HOME

### Type 1 - evaluation :

Type 1 - evaluation Sin 60° * cos 30° + sin 30° * cos60° =√3 / 2 * √3 / 2 + 1 / 2 * 1 / 2 = 3 / 4 + 1 / 4 = 4 / 4 = 1 Next Slide Previous Slide HOME

### Type 2 – finding values of a and b :

Type 2 – finding values of a and b If Tan (A+B) = √3 and tan ( A – B) = 1/ √3 ; 0° < A + B ≤ 90° ; A> B , find A and B. Solution – tan (A + B ) = √3 tan (A+ B ) = tan 60° A+ B = 60° - ( 1) tan (A- B) = 1 / √3 tan (A- B) = tan 30° A – B = 30° - ( 2 ) From ( 1 ) & ( 2) A = 45 ° B = 15 ° Next Slide Previous Slide HOME

### FORMULAS :

FORMULAS Sin ( 90° – θ ) = Cos θ Cos ( 90° – θ ) = Sin θ Tan ( 90° – θ ) = Cot θ Cot ( 90° – θ ) = Tan θ Cosec ( 90° – θ ) = Sec θ Sec ( 90° – θ ) = Cosec θ Next Slide Previous Slide HOME

### Example on formulas :

Example on formulas Evaluate : - (1) Sin 18 ° / Cos 72 ° = Sin (90 – 72 ) ° / Cos 72 ° = Cos 72 ° / Cos 72 ° = 1 ( 2) Cos 48 ° – Sin 42 ° = Cos ( 90 – 42 ) ° – Sin 42 ° = Sin 42 ° – Sin 42 ° = 0 Next Slide Previous Slide HOME

### Main identities :

Main identities Sin²θ + Cos² θ = 1 1 + Tan² θ = Sec² θ 1 + Cot² θ = Cosec² θ Sinθ / Cos θ = Tan θ Cosθ / Sin θ = Cot θ Sin² θ / Cos² θ = Tan² θ Cos² θ / Sin² θ = Cot² θ Next Slide Previous Slide HOME

### Steps of proving the identities :

Steps of proving the identities Solve the left hand side or right hand side of the identity. Use an identity if required. Use formulas if required. Convert the terms in the form of sinθ or cos θ according to the question. Divide or multiply the L.H.S. by sin θ or cos θ if required. Then solve the R.H.S. if required. Lastly , verify that if L.H.S. = R.H.S. Next Slide Previous Slide HOME

### THE END :

THE END PROJECT MADE BY – SHUBHAM KUMAR Previous Slide HOME